220=(x+5)(2x+12)

Solution for 220=(x+5)(2x+12) equation:

220=(x+5)(2x+12)

We move all terms to the left:

220-((x+5)(2x+12))=0

We multiply parentheses ..

-((+2x^2+12x+10x+60))+220=0


We calculate terms in parentheses: -((+2x^2+12x+10x+60)), so:

(+2x^2+12x+10x+60)

We get rid of parentheses

2x^2+12x+10x+60

We add all the numbers together, and all the variables

2x^2+22x+60

Back to the equation:

-(2x^2+22x+60)


We get rid of parentheses

-2x^2-22x-60+220=0

We add all the numbers together, and all the variables

-2x^2-22x+160=0

a = -2; b = -22; c = +160;
Δ = b2-4ac
Δ = -222-4·(-2)·160
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-42}{2*-2}=\frac{-20}{-4}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+42}{2*-2}=\frac{64}{-4}
=-16
$